What is the intuitive explanation of (non-singular part of) a singular elliptic curve being isomorphic to either $K^{*}$ or $K$?

Question

Is there a rough explanation without using explicit computation?

Answer 1

The isomorphism comes about by looking at lines through the singular point $P$. Indeed, for any point $Q \neq P$ on $C$, the unique line in $\mathbb P^2$ through $P$ and $Q$ does not intersect $C$ anywhere else, because it already intersects with multiplicity $3$ (namely, once at $Q$ and twice at $P$). Thus, we get a map $$f \colon C\setminus\{P\} \to \mathbb P^1$$ mapping $Q$ to the direction at $P$ of the line through $Q$. Now what is the image of $f$?

Let $\ell$ be any line through $P$. Then $\ell$ intersects $C$ at $P$ with multiplicity at least $2$. It intersects $C$ with multiplicity $3$ at $P$ exactly if $\ell$ is a tangent direction at $P$ (you can use this as the definition of tangent direction, if you want). Now if $\ell$ intersects $C$ at $P$ with multiplicity $2$, then there must be a third intersection $Q$, so $\ell$ is in the image of $f$. If $\ell$ intersects $C$ with multiplicity $3$ at $P$, then there is no such $Q$, so $\ell$ is not in the image.

We have two cases:

• If $P$ is a node, then there are two tangent directions at $P$. Thus, the image of $f$ is $\mathbb P^1$ minus two points, which is isomorphic to $\mathbb A^1\setminus\{0\}$.
• If $P$ is a cusp, then there is only one tangent direction, so we get $\mathbb P^1$ minus one point. This is isomorphic to $\mathbb A^1$.

Of course, this doesn't say anything about the group structure yet.

Answer 2

Here is an answer that I find particularly appealing. It will require you to believe two facts, the proofs of which I'll sketch.

Theorem: Let $k$ be a field. Then, geometrically, every connected smooth $1$-dimensional linear algebraic group $G/k$ is geometrically isomorphic to a torus or the additive group. In symbols, $G_{\overline{k}}\cong \mathbf{G}_a$ or $G_{\overline{k}}\cong \mathbf{G}_m$.

Proof(Sketch): By setup, we may as well assume that $k=\overline{k}$. If we can show that, as a scheme, $G$ is $\mathbb{A}^1$ or $\mathbb{A}^1-\{0\}$ we're done since, one can check, by hand, that all group structures on these are isomorphic to the obvious ones.

To see that $G$ must be, as a scheme, either $\mathbb{A}^1$ or $\mathbb{A}^1-\{0\}$. We proceed as follows. Consider the unique integral projectivization $X$ of $G$. We claim that this is $\mathbb{P}^1$. Indeed, since $G$ is infinite, it has infinitely many $k$-points which give rise to automorphisms of $G$ as a scheme. These extend, by the valuative criterion, to isomorphisms of $X$. If $g(X)>2$ it's well known that it has only finitely many automorphisms. To eliminate $g(X)=1$, note that the automorphisms of $X$ we obtain by extending the translations of $G$ stabilize, by setup, the finite set $X-G$. Assume, that $g(X)=1$. Then, we can find infinitely many automorphisms of $X$ that fix some point $x_0\in X-G$. Note then that we have produced infinitely many automorphisms of the elliptic curve $(X,x_0)$ which is impossible. Thus, $X=\mathbb{P}^1$.

We're done if we can show that $X-G$ has at most $2$ points. But, note that we can find infinitely many automorphisms of $X$ fixing $X-G$ pointwise (by pigeonhole) and since any automorphism of $\mathbb{P}^1$ fixing three points is trivial, we conclude that $\#(X-G)<3$ as desired. $\blacksquare$

Usin this, we deduce the following:

Theorem: Let $k$ be a perfect field. Then, if $G$ is a $1$-dimensional smooth connected linear algebraic group then, $G$ is one of the following groups:

1. $\mathbf{G}_a$
2. $\mathbf{G}_{m,k}$.
3. $\text{Res}^1_{L/k}\mathbf{G}_{m,L}$ for $L/k$ a degree $2$ Galois extensions(or, I guess since we're perfect, any degree $2$-extension.)

Proof(Sketch): By the previous theorem it suffices to classify twists of $\mathbf{G}_a$ and $\mathbf{G}_m$.

But, the twists of $\mathbf{G}_a$ are in bijection with

$$H^1(G_k,\text{Aut}(\mathbf{G}_{a,\overline{k}}))=H^1(G_k,k)=0$$

and thus there are no non-trivial twists.

Let's calcuate the twists of $\mathbf{G}_m$. These are

$$H^1(G_k,\text{Aut}(\mathbf{G}_{m,\overline{k}})=H^1(G_k,\mathbb{Z}/2\mathbb{Z})=\text{Hom}_{\text{cont.}}(G_k,\mathbb{Z}/2\mathbb{Z})$$

which corresponds to the Galois extensions of $k$ of degree at most $2$. To see that the types 2. and 3. groups are precisely these, note that this tells us there is a unique torus splitting over $L/k$. Since we've listed a torus splitting over $L/k$ for all $L$, we're done. $\blacksquare$

As a corollary of this, we see the following:

Corollary: Let $\mathbb{F}_q$ be the finite field of order $q$, then the complete list of $1$-dimensional connected smooth affine groups is

1. $\mathbf{G}_{a,\mathbb{F}_q}$
2. $\mathbf{G}_{m,\mathbb{F}_q}$
3. $\mathrm{Res}^1_{\mathbb{F}_{q^2}/\mathbb{F}_q}\mathbf{G}_{m,\mathbb{F}_{q^2}}$

Thus, let's suppose that $E$ is an elliptic curve over $K$ the fraction field of a DVR $R$ with residue field $\mathbb{F}_q$. Then, if one takes a model $\mathcal{E}/R$ and considers the reduction $\mathcal{E}_{\mathbb{F}_q}$ then it's classical that $\mathcal{E}^\text{sm}_{\mathbb{F}_q}$ (which differs from $\mathcal{E}_{\mathbb{F}_q}$ by one point) is a smooth connected $1$-dimensional linear algebraic groups. So, it must be one of the three types I listed above.

All of these occur. If it's type 1., we see that $p$ has additive reduction, if it's type $2.$ we say that $p$ is split multiplicative reduction (since we have a split torus!), and if it's type 3. we say it has non-split multiplicative reduction (since it's a non-split torus!).

Hope this helps.